UVA11374最短路模板

蓝书p329

lrj模板

// UVa11374 Airport Express
// Rujia Liu
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;

const int INF = 1000000000;
const int maxn = 500 + 10;

struct Edge {
int from, to, dist;
};

struct HeapNode {
int d, u;
bool operator < (const HeapNode& rhs) const {
return d > rhs.d;
}
};

struct Dijkstra {
int n, m;
vector<Edge> edges;
vector<int> G[maxn];
bool done[maxn]; // 是否已永久标号
int d[maxn]; // s到各个点的距离
int p[maxn]; // 最短路中的上一条弧

void init(int n) {
this->n = n;
for(int i = 0; i < n; i++) G[i].clear();
edges.clear();
}

void AddEdge(int from, int to, int dist) {
edges.push_back((Edge){from, to, dist});
m = edges.size();
G[from].push_back(m-1);
}

void dijkstra(int s) {
priority_queue<HeapNode> Q;
for(int i = 0; i < n; i++) d[i] = INF;
d[s] = 0;
memset(done, 0, sizeof(done));
Q.push((HeapNode){0, s});
while(!Q.empty()) {
HeapNode x = Q.top(); Q.pop();
int u = x.u;
if(done[u]) continue;
done[u] = true;
for(int i = 0; i < G[u].size(); i++) {
Edge& e = edges[G[u][i]];
if(d[e.to] > d[u] + e.dist) {
d[e.to] = d[u] + e.dist;
p[e.to] = G[u][i];
Q.push((HeapNode){d[e.to], e.to});
}
}
}
}

// dist[i]为s到i的距离，paths[i]为s到i的最短路径（经过的结点列表，包括s和t）
void GetShortestPaths(int s, int* dist, vector<int>* paths) {
dijkstra(s);
for(int i = 0; i < n; i++) {
dist[i] = d[i];
paths[i].clear();
int t = i;
paths[i].push_back(t);
while(t != s) {
paths[i].push_back(edges[p[t]].from);
t = edges[p[t]].from;
}
reverse(paths[i].begin(), paths[i].end());
}
}
};

//////// 题目相关
Dijkstra solver;
int d1[maxn], d2[maxn];
vector<int> paths1[maxn], paths2[maxn];

int main() {
int kase = 0, N, S, E, M, K, X, Y, Z;
while(scanf("%d%d%d%d", &N, &S, &E, &M) == 4) {
solver.init(N);

S--; E--; // 编号从0~N-1
for(int i = 0; i < M; i++) {
scanf("%d%d%d", &X, &Y, &Z); X--; Y--;
solver.AddEdge(X, Y, Z);
solver.AddEdge(Y, X, Z);
}
solver.GetShortestPaths(S, d1, paths1); // S到所有点的距离和路径
solver.GetShortestPaths(E, d2, paths2); // T到所有点的距离和路径

int ans = d1[E]; // 初始解解为直达距离
vector<int> path = paths1[E]; // 初始解的station序列
int midpoint = -1; // 不坐商业线

scanf("%d", &K);
for(int i = 0; i < K; i++) {
scanf("%d%d%d", &X, &Y, &Z); X--; Y--;
for(int j = 0; j < 2; j++) { // j=0代表商业线坐X->Y，j=1代表Y->X
if(d1[X] + d2[Y] + Z < ans) {
ans = d1[X] + d2[Y] + Z;
path = paths1[X];
for(int j = paths2[Y].size()-1; j >= 0; j--) // 从Y到T的距离要反过来
path.push_back(paths2[Y][j]);
midpoint = X;
}
swap(X, Y);
}
}

if(kase != 0) printf("\n");
kase++;

for(int i = 0; i < path.size()-1; i++) printf("%d ", path[i]+1);
printf("%d\n", E+1);
if(midpoint == -1) printf("Ticket Not Used\n"); else printf("%d\n", midpoint+1);
printf("%d\n", ans);
}
return 0;
}